9709/22 Oct/Nov 2025 — Mathematics Solved Past Paper

Q14MMedium-EasyLogarithmic and Exponential Functions

Solve the equation $\ln(3x + 5) - \ln(x - 2) = 4$ . Give your answer in an exact form.

DifficultyMedium-Easy

Worked solution

Approach

Use the quotient law of logarithms to combine the two logarithms into one, then exponentiate both sides to remove the logarithm. Solve the resulting linear equation for $x$ and express the answer exactly.

Working

Combine the logarithms using $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$ :

\ln(3x+5) - \ln(x-2) = \ln\left(\frac{3x+5}{x-2}\right) = 4

Exponentiate both sides:

\frac{3x+5}{x-2} = e^4

Multiply through by $x-2$ :

3x+5 = e^4(x-2)

Expand and collect the $x$ -terms:

3x+5 = e^4 x - 2e^4

3x - e^4 x = -2e^4 - 5

(3-e^4)x = -2e^4 - 5

Divide by $3-e^4$ and simplify the signs:

x = \frac{2e^4 + 5}{e^4 - 3}

Since $e^4 - 3 > 0$ and this value is greater than $2$ , the solution is in the domain of the original logarithms.

Answer

x = \frac{2e^4 + 5}{e^4 - 3}

Final answer

x = (2e^4 + 5)/(e^4 - 3)

Detailed explanation

Walkthrough

Start by checking the domain: we need $3x+5>0$ and $x-2>0$ , so $x>2$ . This ensures both logarithms are defined.

The equation has two logarithms subtracted. The quotient law of logarithms says $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$ , so combine the left-hand side into a single logarithm:

\ln\left(\frac{3x+5}{x-2}\right)=4

This is the first mark: applying the logarithm property correctly.

Next, because $\ln$ and $e$ are inverse functions, exponentiate both sides. Raising $e$ to each side removes the logarithm:

\frac{3x+5}{x-2}=e^4

This gives the equation without logarithms and earns the second mark.

Now solve this equation. Multiply both sides by $x-2$ to clear the fraction:

3x+5=e^4(x-2)

Expand the right-hand side and collect the $x$ -terms on one side:

3x+5=e^4x-2e^4

3x-e^4x=-2e^4-5

(3-e^4)x=-2e^4-5

Finally divide by $3-e^4$ . Multiplying top and bottom by $-1$ gives the cleaner form:

x=\frac{2e^4+5}{e^4-3}

This exact value is the final answer. We should check it satisfies the domain. Since $e^4 \approx 54.6$ , $x \approx \frac{109.2+5}{51.6} \approx 2.21$ , which is greater than $2$ , so it is valid.

Key Takeaways

This question tests the quotient law of logarithms, the inverse relationship between $\ln$ and $e$ , and solving a linear equation. It also reminds us to keep answers exact when requested and to be aware of the domain of logarithmic functions.

Common Mistakes

Using $\ln(3x+5)-\ln(x-2)=\ln((3x+5)-(x-2))$ instead of the quotient law. The correct property is $\ln a-\ln b=\ln\left(\frac{a}{b}\right)$ .
Forgetting to exponentiate both sides properly, for example writing $\frac{3x+5}{x-2}=4$ instead of $e^4$ .
Making a sign error when moving terms: $3x-e^4x=-2e^4-5$ , not $2e^4+5$ on the right before simplifying.
Giving a decimal approximation instead of the exact form required.
Forgetting to check the domain, which can lead to accepting an extraneous solution.

Things to Be Careful About

The domain condition $x>2$ is important. If the final answer were less than or equal to $2$ , the original equation would be undefined. The mark scheme requires the equation without logarithms to be correct, and the final answer must be exact. A sign slip in solving is condoned only if the method is otherwise correct, but the final exact form should be $\frac{2e^4+5}{e^4-3}$ .

Techniques used

combine logarithms using the quotient lawexponentiate both sides to remove logarithmssolve the resulting linear equationverify the solution lies in the domain

Approach

The equation involves both $\tan^2\theta$ and $\sec\theta$ . Use the Pythagorean identity $\sec^2\theta = 1 + \tan^2\theta$ to rewrite $\tan^2\theta$ in terms of $\sec\theta$ , forming a quadratic equation in $\sec\theta$ . Solve the quadratic, convert to $\cos\theta$ , and find all angles in the range $-180^\circ < \theta < 180^\circ$ .

Working

From the identity, $\tan^2\theta = \sec^2\theta - 1$ . Substitute into the equation:

2(\sec^2\theta - 1) + 3\sec\theta = 18

Expand and rearrange:

2\sec^2\theta + 3\sec\theta - 20 = 0

Let $u = \sec\theta$ :

2u^2 + 3u - 20 = 0

Factorise:

(2u - 5)(u + 4) = 0

So $u = \frac{5}{2}$ or $u = -4$ .

Case 1: $\sec\theta = \frac{5}{2}$ , so $\cos\theta = \frac{2}{5}$ .

\theta = \pm\cos^{-1}\left(\frac{2}{5}\right) = \pm 66.42^\circ

Case 2: $\sec\theta = -4$ , so $\cos\theta = -\frac{1}{4}$ .

\theta = \pm\cos^{-1}\left(-\frac{1}{4}\right) = \pm 104.48^\circ

All four values lie within the interval $-180^\circ < \theta < 180^\circ$ .

Answer

\theta = \pm 66.42^\circ, \pm 104.48^\circ

Final answer

θ = ±66.42°, ±104.48°

Detailed explanation

Walkthrough

We need to solve $2\tan^2\theta + 3\sec\theta = 18$ for $-180^\circ < \theta < 180^\circ$ . The equation mixes two trigonometric functions, $\tan\theta$ and $\sec\theta$ . To solve it, we want to express everything in terms of a single function. The Pythagorean identity $\sec^2\theta = 1 + \tan^2\theta$ links these two functions, so rearranging it to $\tan^2\theta = \sec^2\theta - 1$ lets us replace every $\tan^2\theta$ with an expression in $\sec\theta$ .

Substituting gives $2(\sec^2\theta - 1) + 3\sec\theta = 18$ , which expands to $2\sec^2\theta - 2 + 3\sec\theta = 18$ and simplifies to $2\sec^2\theta + 3\sec\theta - 20 = 0$ . This is a quadratic equation in $\sec\theta$ . Setting $u = \sec\theta$ makes it easier to solve: $2u^2 + 3u - 20 = 0$ .

Factorising, $(2u - 5)(u + 4) = 0$ , so $u = \frac{5}{2}$ or $u = -4$ . Converting back with $\sec\theta = \frac{1}{\cos\theta}$ , we get $\cos\theta = \frac{2}{5}$ or $\cos\theta = -\frac{1}{4}$ .

For $\cos\theta = \frac{2}{5}$ , the principal value is $\cos^{-1}\left(\frac{2}{5}\right) = 66.42^\circ$ . Because cosine is even, $\cos(-\theta) = \cos\theta$ , so $-66.42^\circ$ is also a solution. Both lie in the interval.

For $\cos\theta = -\frac{1}{4}$ , the principal value is $\cos^{-1}\left(-\frac{1}{4}\right) = 104.48^\circ$ , and by evenness $-104.48^\circ$ is also a solution. Both lie in the interval.

Thus the four solutions are $\theta = \pm 66.42^\circ, \pm 104.48^\circ$ .

Key Takeaways

The Pythagorean identity $\sec^2\theta = 1 + \tan^2\theta$ (equivalently $\tan^2\theta = \sec^2\theta - 1$ ) lets you rewrite a trigonometric equation in terms of a single function.
A quadratic in $\sec\theta$ is solved by factorising, then converted back using $\sec\theta = \frac{1}{\cos\theta}$ .
The cosine function is even, so a solution $\theta$ in $(0^\circ, 180^\circ)$ always has a matching negative solution $-\theta$ in $(-180^\circ, 0^\circ)$ .

Common Mistakes

Using the identity incorrectly, e.g. writing $\tan^2\theta = \sec^2\theta + 1$ instead of $\sec^2\theta - 1$ .
Forgetting to distribute the factor of 2 when substituting $2(\sec^2\theta - 1)$ .
Only finding the positive angles and omitting the negative ones.
Not checking that all solutions lie within the given interval.

Things to Be Careful About

The interval is $-180^\circ < \theta < 180^\circ$ , a strict open interval. All four angles satisfy this.
When $\cos\theta = -\frac{1}{4}$ , the solutions lie in the second and third quadrants; since the interval is symmetric about $0^\circ$ , they are written as $\pm 104.48^\circ$ rather than $104.48^\circ$ and $255.52^\circ$ .
The mark scheme requires the identity step (M1), the correct quadratic (A1), solving the quadratic (DM1), and then the angles — show each step to earn full marks.

Techniques used

apply the Pythagorean identity to express tan^2 in terms of sec^2form a quadratic equation in sec thetafactorise the quadratic to find values of sec thetaconvert sec to cos and apply inverse cosineuse the evenness of cosine to find all angles in the given range

Approach

Since both sides of the inequality are non-negative, square both sides to remove the modulus signs and solve the resulting quadratic inequality.

Working

(3x-4)^2 \leq (2x+5)^2

Expand both sides:

9x^2 - 24x + 16 \leq 4x^2 + 20x + 25

Bring all terms to the left-hand side:

5x^2 - 44x - 9 \leq 0

Factorise:

(5x+1)(x-9) \leq 0

The critical values are $x = -\frac{1}{5}$ and $x = 9$ . Since the quadratic has a positive leading coefficient, it is negative or zero between these roots:

-\frac{1}{5} \leq x \leq 9

Answer

-\frac{1}{5} \leq x \leq 9

Final answer

-1/5 ≤ x ≤ 9

Detailed explanation

Walkthrough

We need to solve $|3x-4| \leq |2x+5|$ . Because both sides are absolute values, they are never negative, so squaring both sides does not change the inequality. Squaring removes the modulus signs and gives $(3x-4)^2 \leq (2x+5)^2$ .

Expanding gives $9x^2 - 24x + 16 \leq 4x^2 + 20x + 25$ . Moving everything to one side gives $5x^2 - 44x - 9 \leq 0$ . This quadratic factorises as $(5x+1)(x-9)$ , so the critical values are $x = -\frac{1}{5}$ and $x = 9$ .

Since the coefficient of $x^2$ is positive, the quadratic is negative between its roots and positive outside them. Therefore the solution is $-\frac{1}{5} \leq x \leq 9$ .

Key Takeaways

This question tests how to handle a modulus inequality. Squaring is valid because both sides are non-negative. It also tests solving a quadratic inequality by factorising and interpreting the sign of the quadratic.

Common Mistakes

Forgetting that squaring both sides is only valid when both sides are non-negative; here it is valid because absolute values are always non-negative.
Making a sign error when expanding $(3x-4)^2$ or $(2x+5)^2$ .
Misidentifying the interval: for a positive quadratic, the inequality $\leq 0$ holds between the roots, not outside them.
Using strict inequalities instead of $\leq$ at the endpoints.

Things to Be Careful About

The endpoints $x = -\frac{1}{5}$ and $x = 9$ must be included because the original inequality allows equality. The final answer must use $\leq$ , not $<$ . Also, when factorising $5x^2 - 44x - 9$ , check that $(5x+1)(x-9)$ expands correctly.

Techniques used

square both sides of a modulus inequalityexpand and rearrange into a quadratic inequalityfactorise a quadratic expressioninterpret the sign of a quadratic inequality

(b)

Hence find the largest integer $N$ satisfying the inequality $|3 \times 7^{0.01N} - 4| \leq |2 \times 7^{0.01N} + 5|$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Let $x = 7^{0.01N}$ . Use the result from part (a): the modulus inequality is equivalent to $-\frac{1}{5} \leq x \leq 9$ . Since $7^{0.01N}$ is always positive, the lower bound is automatically satisfied, so only the upper bound needs to be solved with logarithms.

Working

From part (a),

-\frac{1}{5} \leq 7^{0.01N} \leq 9

Since $7^{0.01N} > 0$ , the left inequality is automatic. Therefore:

7^{0.01N} \leq 9

Take logarithms of both sides:

0.01N \ln 7 \leq \ln 9

N \leq \frac{100 \ln 9}{\ln 7}

Evaluate:

N \leq 112.915\ldots

The largest integer satisfying this is:

N = 112

Answer

112

Final answer

112

Detailed explanation

Walkthrough

Part (a) showed that $|3x-4| \leq |2x+5|$ is equivalent to $-\frac{1}{5} \leq x \leq 9$ . In part (b), the expression $7^{0.01N}$ plays the role of $x$ .

So we need $-\frac{1}{5} \leq 7^{0.01N} \leq 9$ . Since $7^{0.01N}$ is always positive, the lower bound is automatically true. The only real restriction is $7^{0.01N} \leq 9$ .

Take natural logarithms: $0.01N \ln 7 \leq \ln 9$ . Dividing by the positive $0.01 \ln 7$ gives $N \leq \frac{100\ln 9}{\ln 7}$ . Evaluating gives approximately $112.915$ , so the largest integer is $112$ .

Key Takeaways

This part shows how to use logarithms to solve an inequality with the unknown in the exponent. It also demonstrates the importance of using the result from a previous part: the modulus inequality is translated directly into a simple exponential inequality.

Common Mistakes

Forgetting the lower bound from part (a); here it is automatic because the exponential is positive, but it should still be recognised.
Rounding $112.915$ up to $113$ instead of down to $112$ .
Using base 10 logarithms inconsistently; any base works as long as it is used on both sides.
Dividing by a negative quantity without reversing the inequality; here $\ln 7$ and $0.01$ are positive, so no reversal is needed.

Things to Be Careful About

The question asks for the largest integer $N$ , so after obtaining $N \leq 112.915\ldots$ , choose $112$ , not $113$ . If using trial and improvement, you must check both $N=112$ and $N=113$ : $7^{1.12}<9$ and $7^{1.13}>9$ .

Techniques used

substitute an exponential expression into a solved inequalitytake logarithms of both sidesdivide by the logarithm coefficientround down to the largest integer

Q4Medium-EasyAlgebra

The polynomial $p(x)$ is defined by

p(x) = x^4 - 10x^3 + 20x^2 - 30x + 40.

(a)

Find the quotient when $p(x)$ is divided by $(x^2 + 3)$ and show that the remainder is $-11$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use polynomial long division to divide $p(x)$ by $x^2+3$ . Because the divisor is quadratic, the quotient will be quadratic; continue until the remainder has degree less than 2.

Working

Divide the leading term:

x^4 \div x^2 = x^2

Multiply and subtract:

x^4 - 10x^3 + 20x^2 - 30x + 40 - x^2(x^2+3) = -10x^3 + 17x^2 - 30x + 40

Divide the new leading term:

-10x^3 \div x^2 = -10x

Multiply and subtract:

-10x^3 + 17x^2 - 30x + 40 - (-10x)(x^2+3) = 17x^2 + 40

Divide the new leading term:

17x^2 \div x^2 = 17

Multiply and subtract:

17x^2 + 40 - 17(x^2+3) = 40 - 51 = -11

Therefore the quotient is $x^2 - 10x + 17$ and the remainder is $-11$ .

Answer

\text{Quotient} = x^2 - 10x + 17, \qquad \text{Remainder} = -11

Final answer

Quotient = x^2 - 10x + 17; remainder = -11

Detailed explanation

Walkthrough

We need to divide $p(x)$ by $x^2+3$ . Since the divisor is quadratic, the quotient will be quadratic. Start with the highest power: $x^4 \div x^2 = x^2$ . Multiply the divisor by $x^2$ and subtract from $p(x)$ ; this removes the $x^4$ term and leaves a cubic. Then divide the new leading term $-10x^3$ by $x^2$ to get $-10x$ , multiply and subtract again. Finally divide $17x^2$ by $x^2$ to get $17$ , multiply and subtract. The final leftover is $-11$ , so the quotient is $x^2-10x+17$ and the remainder is $-11$ . This confirms the result.

Key Takeaways

Polynomial division by a quadratic follows the same pattern as numeric long division: divide leading terms, multiply, subtract, repeat. The identity $p(x) = (x^2+3)Q(x) + R$ is useful; here $R=-11$ .

Common Mistakes

Stopping too early: must continue until the degree of the remainder is less than the degree of the divisor.
Sign errors when subtracting, especially with $-10x(x^2+3) = -10x^3 - 30x$ .
For "show that" questions, an unsupported statement of the remainder may not earn the final mark; show the division or multiplication.

Things to Be Careful About

The divisor is $x^2+3$ , not $x^2-3$ ; keep signs correct.
When subtracting, subtract the whole product, not just the leading term.
The mark scheme requires necessary detail to confirm the remainder is $-11$ .

Techniques used

perform polynomial long division by a quadratic divisormultiply the divisor by each quotient termsubtract to find the remainder

(b)

Hence find the real roots of the equation $p(x) + 11 = 0$ . Give your answers in exact form.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

From part (a), write $p(x)+11$ in factorised form using the quotient and divisor. Then solve the resulting equation, remembering that only real roots are required.

Working

From part (a):

p(x) = (x^2+3)(x^2-10x+17) - 11

Therefore:

p(x)+11 = (x^2+3)(x^2-10x+17) = 0

The factor $x^2+3$ is always positive for real $x$ , so it gives no real roots. Solve the quadratic factor:

x^2 - 10x + 17 = 0

Using the quadratic formula:

x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(17)}}{2(1)}

x = \frac{10 \pm \sqrt{100-68}}{2} = \frac{10 \pm \sqrt{32}}{2}

x = \frac{10 \pm 4\sqrt{2}}{2} = 5 \pm 2\sqrt{2}

These are the only real roots.

Answer

x = 5 + 2\sqrt{2} \quad \text{or} \quad x = 5 - 2\sqrt{2}

Final answer

x = 5 + 2√2, x = 5 - 2√2

Detailed explanation

Walkthrough

From part (a), $p(x) = (x^2+3)(x^2-10x+17)-11$ . Therefore $p(x)+11 = (x^2+3)(x^2-10x+17)$ . Setting this equal to zero gives two factors. The factor $x^2+3$ is always positive for real $x$ , so it contributes no real roots. Solve $x^2-10x+17=0$ using the quadratic formula. The discriminant is $100-68=32$ , so $\sqrt{32}=4\sqrt{2}$ , and $x=(10 \pm 4\sqrt{2})/2 = 5 \pm 2\sqrt{2}$ . These are the only real roots.

Key Takeaways

A factorised polynomial equation can be solved by setting each factor to zero. A quadratic with positive leading coefficient and no real roots, such as $x^2+3$ , can be ignored when only real roots are required. Exact surd answers should be simplified.

Common Mistakes

Trying to solve the quartic directly instead of using the factorisation.
Forgetting that $x^2+3=0$ has no real solutions.
Giving $16 \pm 2\sqrt{2}$ or other unsimplified or incorrect forms; the mark scheme requires $5 \pm \sqrt{8}$ or $5 \pm 2\sqrt{2}$ .
Not simplifying $\sqrt{32}$ to $4\sqrt{2}$ before dividing by 2.

Things to Be Careful About

The mark scheme says "Do not ISW" (ignore subsequent working), so an incorrect final form cannot be rescued by later correct work.
"Exact form" means leave surds, not decimals.
Since $x^2+3>0$ for all real $x$ , there are exactly two real roots, not four.

Techniques used

use the quotient from polynomial division to factorise the quarticsolve a quadratic equation using the quadratic formulaidentify and discard a factor with no real roots

Q5MediumTrigonometryDifferentiationIntegration

Fig. 1

The diagram shows the curve with equation $y = 4\cos 2x + 8\sin x$ for $0 \leq x \leq \pi$ . The maximum points on the curve are denoted by $A$ and $B$ , and the shaded region is bounded by the line segment $AB$ and the curve.

(a)

Find the coordinates of $A$ and $B$ .

5M Similar

DifficultyMedium

Worked solution

Approach

To find the maximum points on the curve, differentiate $y$ with respect to $x$ , set $\frac{dy}{dx} = 0$ , and solve the resulting trigonometric equation using the double angle identity $\sin 2x = 2\sin x \cos x$ . Then verify which solutions correspond to maximum points by checking the $y$ -values or the second derivative.

Working

Given:

y = 4\cos 2x + 8\sin x

Differentiate with respect to $x$ :

\frac{dy}{dx} = -8\sin 2x + 8\cos x

Set $\frac{dy}{dx} = 0$ :

-8\sin 2x + 8\cos x = 0

Apply the double angle identity $\sin 2x = 2\sin x \cos x$ :

-8(2\sin x \cos x) + 8\cos x = 0

-16\sin x \cos x + 8\cos x = 0

Factorise:

8\cos x(1 - 2\sin x) = 0

This gives two cases:

\cos x = 0 \quad \text{or} \quad 1 - 2\sin x = 0

For $0 \leq x \leq \pi$ :

$\cos x = 0 \implies x = \frac{\pi}{2}$
$\sin x = \frac{1}{2} \implies x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$

Now find the $y$ -coordinates at each critical point:

At $x = \frac{\pi}{6}$ :

y = 4\cos\left(\frac{\pi}{3}\right) + 8\sin\left(\frac{\pi}{6}\right) = 4\left(\frac{1}{2}\right) + 8\left(\frac{1}{2}\right) = 2 + 4 = 6

At $x = \frac{\pi}{2}$ :

y = 4\cos(\pi) + 8\sin\left(\frac{\pi}{2}\right) = 4(-1) + 8(1) = -4 + 8 = 4

At $x = \frac{5\pi}{6}$ :

y = 4\cos\left(\frac{5\pi}{3}\right) + 8\sin\left(\frac{5\pi}{6}\right) = 4\left(\frac{1}{2}\right) + 8\left(\frac{1}{2}\right) = 2 + 4 = 6

Since $y = 6 > 4$ at $x = \frac{\pi}{2}$ , the points $\left(\frac{\pi}{6}, 6\right)$ and $\left(\frac{5\pi}{6}, 6\right)$ are maximum points, and $\left(\frac{\pi}{2}, 4\right)$ is a local minimum.

Therefore:

A = \left(\frac{\pi}{6}, 6\right) \quad \text{and} \quad B = \left(\frac{5\pi}{6}, 6\right)

Answer

A = \left(\frac{\pi}{6}, 6\right), \quad B = \left(\frac{5\pi}{6}, 6\right)

Final answer

A = (π/6, 6) and B = (5π/6, 6)

Detailed explanation

Walkthrough

Step 1: Differentiate the curve equation.
We are given $y = 4\cos 2x + 8\sin x$ . To find stationary points, we need $\frac{dy}{dx}$ . Using the standard derivatives of $\cos$ and $\sin$ along with the chain rule for $\cos 2x$ :

\frac{dy}{dx} = -8\sin 2x + 8\cos x

Step 2: Set the derivative to zero and apply a double angle identity.
Setting $\frac{dy}{dx} = 0$ gives $-8\sin 2x + 8\cos x = 0$ . The term $\sin 2x$ prevents direct solving, so we use the double angle identity $\sin 2x = 2\sin x \cos x$ to express everything in terms of $x$ :

-16\sin x \cos x + 8\cos x = 0

Step 3: Factorise and solve.
Factoring out $8\cos x$ gives $8\cos x(1 - 2\sin x) = 0$ . This splits into two simple equations: $\cos x = 0$ and $\sin x = \frac{1}{2}$ . Within $0 \leq x \leq \pi$ , these yield $x = \frac{\pi}{2}$ , $x = \frac{\pi}{6}$ , and $x = \frac{5\pi}{6}$ .

Step 4: Find $y$ -coordinates and identify maxima.
Substituting each $x$ -value back into the original equation gives $y = 6$ , $y = 4$ , and $y = 6$ respectively. The points with $y = 6$ are the maxima (A and B), while $x = \frac{\pi}{2}$ gives a local minimum.

Key Takeaways

Differentiating trigonometric functions requires the chain rule and standard derivative formulas.
When a double angle term appears in a derivative, applying $\sin 2x = 2\sin x \cos x$ often allows factorisation.
Always check all critical points and substitute back to determine which are maxima versus minima.

Common Mistakes

Forgetting the chain rule when differentiating $\cos 2x$ , giving $-\sin 2x$ instead of $-2\sin 2x$ .
Not using the double angle identity and attempting to solve $-8\sin 2x + 8\cos x = 0$ directly without converting to a single angle.
Finding all critical points but failing to distinguish maxima from the local minimum at $x = \frac{\pi}{2}$ .
Using degrees instead of radians for the angles.

Things to Be Careful About

The domain is $0 \leq x \leq \pi$ , so only solutions within this range are valid.
Angles must be expressed in radians as specified in the mark scheme.
All three critical points ( $x = \frac{\pi}{6}$ , $\frac{\pi}{2}$ , $\frac{5\pi}{6}$ ) must be evaluated; omitting $x = \frac{\pi}{2}$ and incorrectly claiming the others are the only stationary points loses marks.

Techniques used

differentiate trigonometric expressionapply double angle identitysolve trigonometric equationcheck second derivative or sign change to confirm maxima

(b)

Find the exact area of the shaded region.

6M Similar

DifficultyMedium

Worked solution

Approach

The shaded region is bounded above by the horizontal line segment $AB$ (at $y = 6$ ) and below by the curve $y = 4\cos 2x + 8\sin x$ , between $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ . The area can be found by computing the area of the rectangle with width $\frac{5\pi}{6} - \frac{\pi}{6}$ and height $6$ , then subtracting the area under the curve between these $x$ -limits.

Working

Area under the curve from $x = \frac{\pi}{6}$ to $x = \frac{5\pi}{6}$ :

\int_{\pi/6}^{5\pi/6} (4\cos 2x + 8\sin x) \, dx

Integrate term by term:

\left[ 2\sin 2x - 8\cos x \right]_{\pi/6}^{5\pi/6}

Evaluate at the upper limit $x = \frac{5\pi}{6}$ :

2\sin\left(\frac{5\pi}{3}\right) - 8\cos\left(\frac{5\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2}\right) - 8\left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} + 4\sqrt{3} = 3\sqrt{3}

Evaluate at the lower limit $x = \frac{\pi}{6}$ :

2\sin\left(\frac{\pi}{3}\right) - 8\cos\left(\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2}\right) - 8\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} - 4\sqrt{3} = -3\sqrt{3}

Area under the curve:

3\sqrt{3} - (-3\sqrt{3}) = 6\sqrt{3}

Area of the rectangle bounded by $AB$ and the $x$ -axis:

The width is $\frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$ and the height is $6$ :

\text{Rectangle area} = 6 \times \frac{2\pi}{3} = 4\pi

Shaded area:

\text{Shaded area} = 4\pi - 6\sqrt{3}

Answer

4\pi - 6\sqrt{3}

Final answer

4π - 6√3

Detailed explanation

Walkthrough

Step 1: Set up the integral for the area under the curve.
The curve runs from $A$ at $x = \frac{\pi}{6}$ to $B$ at $x = \frac{5\pi}{6}$ . The area under the curve in this interval is:

\int_{\pi/6}^{5\pi/6} (4\cos 2x + 8\sin x) \, dx

Step 2: Integrate.
Using standard integration rules:

$\int 4\cos 2x \, dx = 2\sin 2x$
$\int 8\sin x \, dx = -8\cos x$

So the antiderivative is $2\sin 2x - 8\cos x$ .

Step 3: Evaluate at the limits.
At $x = \frac{5\pi}{6}$ : $\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}$ and $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$ , giving $3\sqrt{3}$ .
At $x = \frac{\pi}{6}$ : $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ , giving $-3\sqrt{3}$ .

The area under the curve is $3\sqrt{3} - (-3\sqrt{3}) = 6\sqrt{3}$ .

Step 4: Find the shaded area by subtraction.
The shaded region lies between the horizontal line $y = 6$ (the line segment $AB$ ) and the curve. The rectangle from $x = \frac{\pi}{6}$ to $x = \frac{5\pi}{6}$ with height $6$ has area $4\pi$ . Subtracting the area under the curve:

\text{Shaded area} = 4\pi - 6\sqrt{3}

Key Takeaways

Integration of $\cos(ax)$ gives $\frac{1}{a}\sin(ax)$ and integration of $\sin(x)$ gives $-\cos(x)$ .
When finding the area between a curve and a horizontal line, compute the rectangle area and subtract the area under the curve.
Careful evaluation of trigonometric values at non-standard angles (like $\frac{5\pi}{3}$ and $\frac{5\pi}{6}$ ) is essential.

Common Mistakes

Incorrect integration signs: $\int \sin x \, dx = -\cos x$ , not $+\cos x$ .
Sign errors when evaluating $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$ (second quadrant, cosine is negative).
Forgetting to subtract the area under the curve from the rectangle area, giving $6\sqrt{3}$ as the final answer instead of $4\pi - 6\sqrt{3}$ .
Using degrees instead of radians when evaluating trigonometric functions in the definite integral.

Things to Be Careful About

The curve dips below $y = 6$ between $A$ and $B$ , so the shaded area is the rectangle minus the area under the curve, not the other way around.
Both $y$ -coordinates of $A$ and $B$ must be equal (both are $6$ ) for the horizontal line segment to form a proper rectangle with the $x$ -axis.
The final answer $4\pi - 6\sqrt{3}$ is positive since $4\pi \approx 12.57$ and $6\sqrt{3} \approx 10.39$ , confirming the area is valid.

Techniques used

integrate trigonometric expressionevaluate definite integral between limitscalculate area of rectanglesubtract area under curve from rectangle area

Q6MediumIntegrationLogarithmic and Exponential FunctionsNumerical Solution of Equations

(a)

Given that $\int_{-2a}^{a} \left( \frac{1}{2}e^{2x} + \frac{1}{4}e^{-x} \right) dx = 5$ , where $a$ is a positive constant, show that

a = \frac{1}{2}\ln\left( 10 + \frac{1}{2}e^{-a} + \frac{1}{2}e^{-4a} \right).

4M Similar

DifficultyMedium

Worked solution

Approach

Integrate the exponential terms term by term, evaluate the definite integral between the limits $-2a$ and $a$ , set the result equal to 5, then rearrange to isolate $a$ using the inverse relationship between $\ln$ and $e^x$ .

Working

Integrate each term:

\int \left( \frac{1}{2}e^{2x} + \frac{1}{4}e^{-x} \right) dx = \frac{1}{4}e^{2x} - \frac{1}{4}e^{-x} + C

Apply the limits $-2a$ and $a$ :

\left[ \frac{1}{4}e^{2x} - \frac{1}{4}e^{-x} \right]_{-2a}^{a} = \left( \frac{1}{4}e^{2a} - \frac{1}{4}e^{-a} \right) - \left( \frac{1}{4}e^{-4a} - \frac{1}{4}e^{2a} \right)

= \frac{1}{4}e^{2a} - \frac{1}{4}e^{-a} - \frac{1}{4}e^{-4a} + \frac{1}{4}e^{2a}

= \frac{1}{2}e^{2a} - \frac{1}{4}e^{-a} - \frac{1}{4}e^{-4a}

Set this equal to 5:

\frac{1}{2}e^{2a} - \frac{1}{4}e^{-a} - \frac{1}{4}e^{-4a} = 5

\frac{1}{2}e^{2a} = 5 + \frac{1}{4}e^{-a} + \frac{1}{4}e^{-4a}

e^{2a} = 10 + \frac{1}{2}e^{-a} + \frac{1}{2}e^{-4a}

Take natural logarithms of both sides:

2a = \ln\left( 10 + \frac{1}{2}e^{-a} + \frac{1}{2}e^{-4a} \right)

a = \frac{1}{2}\ln\left( 10 + \frac{1}{2}e^{-a} + \frac{1}{2}e^{-4a} \right)

Answer

a = \frac{1}{2}\ln\left( 10 + \frac{1}{2}e^{-a} + \frac{1}{2}e^{-4a} \right)

Final answer

a = (1/2) ln(10 + (1/2)e^(-a) + (1/2)e^(-4a))

Detailed explanation

Walkthrough

We are told that the definite integral of $\frac{1}{2}e^{2x} + \frac{1}{4}e^{-x}$ from $-2a$ to $a$ equals 5, and we must show this leads to the given formula for $a$ .

First, integrate each term separately. Recall that $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ . For $\frac{1}{2}e^{2x}$ , the factor from integration is $\frac{1}{2}$ , so we get $\frac{1}{2} \times \frac{1}{2} e^{2x} = \frac{1}{4}e^{2x}$ . For $\frac{1}{4}e^{-x}$ , integration gives $\frac{1}{4} \times \frac{1}{-1} e^{-x} = -\frac{1}{4}e^{-x}$ .

Next, evaluate the antiderivative at the upper limit $a$ and the lower limit $-2a$ , and subtract. At the upper limit we get $\frac{1}{4}e^{2a} - \frac{1}{4}e^{-a}$ . At the lower limit we get $\frac{1}{4}e^{-4a} - \frac{1}{4}e^{2a}$ , because $e^{2(-2a)} = e^{-4a}$ and $e^{-(-2a)} = e^{2a}$ . Subtracting, we must be careful with the minus sign: $F(a) - F(-2a)$ .

Combining like terms gives $\frac{1}{2}e^{2a} - \frac{1}{4}e^{-a} - \frac{1}{4}e^{-4a}$ . Setting this equal to 5 and rearranging, we isolate $\frac{1}{2}e^{2a}$ , then multiply by 2 to get $e^{2a} = 10 + \frac{1}{2}e^{-a} + \frac{1}{2}e^{-4a}$ . Finally, since $\ln$ and $e^x$ are inverse functions, taking natural logs gives $2a = \ln(\ldots)$ , and dividing by 2 gives the required expression.

Key Takeaways

This question tests integration of exponential functions of the form $e^{kx}$ , the careful evaluation of definite integrals with limits involving a parameter, and the use of logarithms to solve for an unknown in an exponent. Understanding that $\ln(e^x) = x$ is essential for the final step.

Common Mistakes

Writing $\int \frac{1}{2}e^{2x}\,dx = \frac{1}{2}e^{2x}$ instead of $\frac{1}{4}e^{2x}$ — forgetting to divide by the coefficient of $x$ in the exponent.
Sign errors when substituting the lower limit, especially for the $-\frac{1}{4}e^{-x}$ term.
Forgetting to distribute the minus sign when computing $F(a) - F(-2a)$ , which would give the wrong combination of terms.
Not dividing by 2 at the end after taking logarithms.

Things to Be Careful About

$e^{2(-2a)} = e^{-4a}$ , not $e^{-2a}$ .
The lower-limit substitution changes the sign of the $e^{2x}$ term: $F(-2a)$ contains $+\frac{1}{4}e^{2a}$ , so subtracting $F(-2a)$ contributes $-\frac{1}{4}e^{2a}$ to the difference, which combines with the $+\frac{1}{4}e^{2a}$ from the upper limit.
Since this is a 'show that' question, all working must be shown — the mark scheme requires the necessary detail.

Techniques used

integrate exponential functions term by termevaluate definite integral at parameter limitsrearrange to isolate the exponential termtake natural logarithms to solve for the exponent

(b)

Hence show by calculation that the value of $a$ lies between $1.0$ and $1.2$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Define the function $f(a) = \frac{1}{2}e^{2a} - \frac{1}{4}e^{-a} - \frac{1}{4}e^{-4a} - 5$ , obtained by rearranging the result of part (a). Evaluate $f$ at $a = 1.0$ and $a = 1.2$ . A change of sign between these values shows that the root $a$ lies between them.

Working

f(a) = \frac{1}{2}e^{2a} - \frac{1}{4}e^{-a} - \frac{1}{4}e^{-4a} - 5

Evaluate at $a = 1.0$ :

f(1.0) = \frac{1}{2}e^{2} - \frac{1}{4}e^{-1} - \frac{1}{4}e^{-4} - 5 \approx 3.69453 - 0.09197 - 0.00458 - 5 = -1.40202\ldots

Evaluate at $a = 1.2$ :

f(1.2) = \frac{1}{2}e^{2.4} - \frac{1}{4}e^{-1.2} - \frac{1}{4}e^{-4.8} - 5 \approx 5.51159 - 0.07530 - 0.00206 - 5 = 0.43423\ldots

Since $f(1.0) < 0$ and $f(1.2) > 0$ , there is a change of sign, so the root $a$ lies between $1.0$ and $1.2$ .

Answer

1.0 < a < 1.2

Final answer

1.0 < a < 1.2

Detailed explanation

Walkthrough

Part (a) established that the value of $a$ satisfies $\frac{1}{2}e^{2a} - \frac{1}{4}e^{-a} - \frac{1}{4}e^{-4a} = 5$ . To locate the root, define $f(a) = \frac{1}{2}e^{2a} - \frac{1}{4}e^{-a} - \frac{1}{4}e^{-4a} - 5$ . The root $a$ is a zero of $f$ .

Evaluate $f$ at the two endpoints $1.0$ and $1.2$ . At $a = 1.0$ , the value is approximately $-1.40$ , which is negative. At $a = 1.2$ , the value is approximately $0.43$ , which is positive. Since $f$ is continuous and changes sign between $1.0$ and $1.2$ , the zero (the value of $a$ ) must lie strictly between them.

Key Takeaways

This part applies the sign-change criterion for locating roots: if a continuous function takes opposite signs at two points, it has at least one root between them. This is the foundational idea behind numerical root-finding.

Common Mistakes

Using the wrong function, e.g. evaluating $\frac{1}{2}\ln(10 + \ldots)$ at the two points without comparing to $a$ .
Arithmetic errors when evaluating the exponentials.
Concluding the result without showing both values and their signs.

Things to Be Careful About

The mark scheme allows either the $f(a)$ form above or comparing $a$ with $\frac{1}{2}\ln(10 + \frac{1}{2}e^{-a} + \frac{1}{2}e^{-4a})$ ; both are acceptable.
You must state that a change of sign justifies the conclusion that $1.0 < a < 1.2$ .

Techniques used

evaluate function at interval endpointsapply sign-change criterion to locate a root

(c)

Use the iterative formula

a_{n+1} = \frac{1}{2}\ln\left( 10 + \frac{1}{2}e^{-a_n} + \frac{1}{2}e^{-4a_n} \right)

to find the value of $a$ correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Apply the iterative formula repeatedly, starting from a value inside the interval found in part (b), such as $a_0 = 1.1$ . Record each iterate to 6 significant figures. Stop when successive iterates agree to the required 4 significant figures.

Working

Start with $a_0 = 1.1$ .

First iteration:

a_1 = \frac{1}{2}\ln\left( 10 + \frac{1}{2}e^{-1.1} + \frac{1}{2}e^{-4.4} \right) = \frac{1}{2}\ln(10.17257\ldots) = 1.15985\ldots

Second iteration:

a_2 = \frac{1}{2}\ln\left( 10 + \frac{1}{2}e^{-1.15985} + \frac{1}{2}e^{-4.6394} \right) = \frac{1}{2}\ln(10.16157\ldots) = 1.15931\ldots

Third iteration:

a_3 = \frac{1}{2}\ln\left( 10 + \frac{1}{2}e^{-1.15931} + \frac{1}{2}e^{-4.63724} \right) = \frac{1}{2}\ln(10.16168\ldots) = 1.15931\ldots

The iterates are $1.15985$ , $1.15931$ , $1.15931$ (each to 6 significant figures). Since $a_2$ and $a_3$ agree to 4 significant figures, the value of $a$ correct to 4 significant figures is $1.159$ .

Answer

a = 1.159

Final answer

a = 1.159

Detailed explanation

Walkthrough

We use the iterative formula $a_{n+1} = \frac{1}{2}\ln\left(10 + \frac{1}{2}e^{-a_n} + \frac{1}{2}e^{-4a_n}\right)$ to refine the approximation of $a$ . A sensible starting value inside the interval $(1.0, 1.2)$ is $a_0 = 1.1$ .

Substituting $a_0 = 1.1$ gives $a_1 = 1.15985\ldots$ . Substituting this into the formula gives $a_2 = 1.15931\ldots$ . One more iteration gives $a_3 = 1.15931\ldots$ . The successive iterates are converging: $1.15985$ , $1.15931$ , $1.15931$ .

To report the answer to 4 significant figures, we compare consecutive iterates. $a_2$ and $a_3$ both round to $1.159$ at 4 significant figures, so we conclude $a = 1.159$ correct to 4 significant figures.

Key Takeaways

This part demonstrates how an iterative formula of the form $x_{n+1} = F(x_n)$ produces a convergent sequence of approximations to a root, and how to decide when to stop iterating: when successive values agree to the required precision.

Common Mistakes

Not showing enough iterations to justify the final answer (the mark scheme requires sufficient iterations or a sign change in $[1.1585, 1.1595]$ ).
Giving the answer to the wrong number of significant figures.
Rounding intermediate values too aggressively, which can change the final digits.

Things to Be Careful About

Each iteration result must be given to 6 significant figures as instructed.
The final answer must be given to exactly 4 significant figures: $1.159$ .
To be safe, iterate until two consecutive values agree to at least 4 significant figures, or verify a sign change in the interval $[1.1585, 1.1595]$ .

Techniques used

apply iterative formula repeatedlytrack convergence of successive approximationsround result to the required significant figures

Q7MediumDifferentiation

A curve has equation $5x^2y + 4e^{2y} - 7x + 10 = 0$ .

(a)

Find an expression in terms of $x$ and $y$ for $\frac{dy}{dx}$ and hence find the gradient of the curve at the point for which $y = 0$ .

6M Similar

DifficultyMedium

Worked solution

Approach

Differentiate both sides of the curve equation implicitly with respect to $x$ , using the product rule for $5x^2y$ and the chain rule for $4e^{2y}$ . Then solve for $\frac{dy}{dx}$ . To find the gradient when $y=0$ , first substitute $y=0$ into the original equation to find the corresponding $x$ -value, then evaluate the derivative.

Working

Differentiate each term. For $5x^2y$ :

\frac{d}{dx}\left(5x^2y\right)=10xy+5x^2\frac{dy}{dx}

For $4e^{2y}$ :

\frac{d}{dx}\left(4e^{2y}\right)=8e^{2y}\frac{dy}{dx}

Differentiating the whole equation:

10xy+5x^2\frac{dy}{dx}+8e^{2y}\frac{dy}{dx}-7=0

Collect the $\frac{dy}{dx}$ terms:

\left(5x^2+8e^{2y}\right)\frac{dy}{dx}=7-10xy

Hence

\frac{dy}{dx}=\frac{7-10xy}{5x^2+8e^{2y}}

When $y=0$ , substitute into the original equation:

5x^2(0)+4e^{0}-7x+10=0

4-7x+10=0 \quad\Rightarrow\quad 14-7x=0

So $x=2$ . Now substitute $x=2$ , $y=0$ into the derivative:

\frac{dy}{dx}=\frac{7-10(2)(0)}{5(2)^2+8e^{0}} =\frac{7}{20+8} =\frac{7}{28} =\frac{1}{4}

Answer

\frac{dy}{dx}=\frac{7-10xy}{5x^2+8e^{2y}},\qquad \text{gradient at }y=0 = \frac{1}{4}

Final answer

dy/dx = (7 - 10xy)/(5x^2 + 8e^(2y)); gradient at y = 0 is 1/4

Detailed explanation

Walkthrough

The curve is defined implicitly, so we differentiate both sides with respect to $x$ rather than trying to rearrange for $y$ first. The term $5x^2y$ is a product of $5x^2$ and $y$ , so it needs the product rule: differentiate $5x^2$ to get $10x$ , keep $y$ , then keep $5x^2$ and multiply by $\frac{dy}{dx}$ because $y$ is a function of $x$ . The term $4e^{2y}$ needs the chain rule: the derivative of $e^{2y}$ with respect to $y$ is $2e^{2y}$ , and multiplying by $\frac{dy}{dx}$ gives $8e^{2y}\frac{dy}{dx}$ . Differentiating $-7x$ gives $-7$ , and the constant $10$ differentiates to $0$ . Collect all $\frac{dy}{dx}$ terms on one side and factor them out. Then divide by the coefficient to get the required expression.

To find the gradient at $y=0$ , first find the corresponding $x$ -value by substituting $y=0$ into the original equation. This gives $4 - 7x + 10 = 0$ , so $x=2$ . Then substitute $x=2$ and $y=0$ into the derivative: the numerator is $7$ , the denominator is $20+8=28$ , so the gradient is $\frac{1}{4}$ .

Key Takeaways

Implicit differentiation is needed when $y$ is not written explicitly in terms of $x$ . Remember to multiply by $\frac{dy}{dx}$ whenever you differentiate a term containing $y$ . The product rule and chain rule often appear together in implicit differentiation. To evaluate a derivative at a point, you may need to find the missing coordinate from the original equation.

Common Mistakes

Forgetting the $\frac{dy}{dx}$ factor when differentiating terms involving $y$ , especially $4e^{2y}$ .
Using the product rule incorrectly on $5x^2y$ ; some candidates differentiate $y$ as if it were $x$ .
Substituting $y=0$ into the derivative without first finding the corresponding $x$ -value.
Sign errors when collecting terms: the derivative equation should give $(5x^2+8e^{2y})\frac{dy}{dx}=7-10xy$ .

Things to Be Careful About

The mark scheme requires evidence of implicit differentiation, so show the $\frac{dy}{dx}$ factors explicitly.
When $y=0$ , $e^{0}=1$ , not $0$ .
The final gradient must be simplified; $\frac{7}{28}$ should be written as $\frac{1}{4}$ .

Techniques used

differentiate implicitly with respect to xapply the product rule to 5x^2yapply the chain rule to 4e^(2y)solve for dy/dxsubstitute y = 0 to find the gradient

(b)

Show that there is no point on the curve at which the tangent is parallel to the $y$ -axis.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

A tangent parallel to the $y$ -axis is vertical, so $\frac{dy}{dx}$ would be undefined. This happens only if the denominator of the derivative expression is zero. Show that the denominator can never be zero.

Working

From part (a),

\frac{dy}{dx}=\frac{7-10xy}{5x^2+8e^{2y}}

For a vertical tangent, the denominator must be zero:

5x^2+8e^{2y}=0

But for all real $x$ and $y$ ,

5x^2 \ge 0 \quad\text{and}\quad 8e^{2y} > 0

Therefore

5x^2+8e^{2y} > 0

for all real $x$ and $y$ . The denominator is never zero, so $\frac{dy}{dx}$ is always defined and the tangent is never parallel to the $y$ -axis.

Answer

There is no point on the curve at which the tangent is parallel to the $y$ -axis, because $5x^2+8e^{2y}>0$ for all real $x$ and $y$ .

Final answer

No such point exists, since 5x^2 + 8e^(2y) > 0 for all real x and y.

Detailed explanation

Walkthrough

A tangent parallel to the $y$ -axis is vertical, meaning its gradient is infinite, i.e. $\frac{dy}{dx}$ is undefined. From the derivative expression, this can only happen when the denominator $5x^2+8e^{2y}$ is zero. We examine this expression. The term $5x^2$ is a square multiplied by a positive constant, so it is always greater than or equal to zero. The term $8e^{2y}$ is always strictly positive because the exponential function is positive for every real input. Therefore their sum is always strictly positive, never zero. Since the denominator can never vanish, the derivative is always finite and no vertical tangent exists.

Key Takeaways

A vertical tangent corresponds to an undefined derivative, which for a quotient expression means the denominator is zero. Exponential functions $e^{2y}$ are always positive, and squares are always non-negative. Combining these facts can prove that an expression is never zero.

Common Mistakes

Trying to set the numerator equal to zero instead of the denominator; that would find horizontal tangents, not vertical ones.
Forgetting that $e^{2y}$ is strictly positive for all real $y$ .
Assuming $x=0$ could make the denominator zero; even if $x=0$ , the term $8e^{2y}$ remains positive.

Things to Be Careful About

The mark scheme says the argument must use the denominator of the derivative, not the derivative evaluated at $y=0$ . The statement must hold for all points on the curve.
If a candidate had a different but equivalent denominator of the form $k_1x^2+k_2e^{2y}$ with $k_1k_2>0$ , the same positivity argument applies.

Techniques used

set the denominator of dy/dx to zerouse positivity of squares and exponentialsconclude no vertical tangent exists

Mathematics 9709/22 — October/November 2025

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